Dijkstra’s Algorithm

Hello, people…! In this post, I will talk about one of the fastest single source shortest path algorithms, which is, the Dijkstra’s Algorithm. The Dijkstra’s Algorithm works on a weighted graph with non-negative edge weights and gives a Shortest Path Tree. It is a greedy algorithm, which sort of mimics the working of breadth first search and depth first search.

The Dijkstra’s Algorithm starts with a source vertex ‘s‘ and explores the whole graph. We will use the following elements to compute the shortest paths –

Priority Queue Q.

An array D, which keeps the record of the total distance from starting vertex s to all other vertices.

Just like the other graph search algorithms, Dijkstra’s Algorithm is best understood by listing out the algorithm in a step-by-step process –

The Initialisation –

D[s], which is the shortest distance to s is set to 0. The distance from the source to itself is 0.

For all the other vertices V, D[V] is set to infinity as we do not have a path yet to them, so we simply say that the distance to them is infinity.

The Priority Queue Q, is constructed which is initially holds all the vertices of the graph. Each vertex V will have the priority D[V].

The Algorithm –

Now, pick up the minimum priority (D[V]) element from Q (which removes it from Q). For the first time, this operation would obviously give s.

For all the vertices, v, adjacent to s, i.e., check if the edge from s → v gives a shorter path. This is done by checking the following condition –

Now pick the next minimum priority element from Q, and repeat the process until there are no elements left in Q.

Let us understand this with the help of an example. Consider the graph below –

Firstly, initialize your components, the shortest distances array D, the priority queue Q. The distance from the source to itself is zero. So, D[s] = 0, and the rest of the array is ∞. The set of vertices v is inserted into the priority queue Q, with a priority D[v]. Now, we start our algorithm by extracting the minimum element from the priority queue.

The minimum element in the priority queue will definitely be s (which is A here). Look at all the adjacent vertices of A. Vertices B, C, D are adjacent to A. We can go to B travelling the edge of weight 2, to C travelling an edge of weight 1, to D travelling an edge of weight 5. The values of D[B], D[C], D[D] are ∞ . We have found a new way of reaching them in 2, 1, 5 units respectively, which is less than ∞, hence a shorter path. This is what the if-condition mentioned above does. So, we update the values of D[B], D[C], D[D] and the priorities of B, C, D, in the priority queue. With this, we have finished processing the vertex A.

Now, the process continues to its next iteration and we extract the minimum element from the priority queue. The minimum element would be vertex C which would be having a priority of 1. Now, look at all the adjacent vertices to C. There’s vertex D. From C, it would take 1 unit of distance to reach D. But to reach C in prior, you need 1 more unit of distance. So, if you go to D, via C, the total distance would be 2 units, which is less than the current value of shortest distance discovered to D, D[D] = 5. So, we reduce the value of D[D] to 2. This reduction is also called as “Relaxation“. With that, we’re done with vertex C.

Now, the process continues to its next iteration and we extract the minimum element from the priority queue. Now, there are two minimum elements, B and D. You can go for anyone. We will go for vertex D. From D, you can go to E and F, with a total distance of 2 + 2 {D[D] + (weight of D → E)}, and 2 + 3. Which is less than ∞, so D[E] becomes 4 and D[F] becomes 5. We’re done with vertex D.

Now, the process continues to its next iteration and we extract the minimum element from the priority queue. The minimum element in the priority queue is vertex B. From vertex B, you can reach F in 2 + 1 units of distance, which is less than the current value of D[F], 5. So, we relax D(F) to 3. From vertex B, you can reach vertex D in 2 + 2 units of distance, which is more than the current value of D(D), 2. This route is not considered as it is clearly proven to be a longer route. With that, we’re done with vertex B.

Now, the process continues to its next iteration and we extract the minimum element from the priority queue. The minimum element in the priority queue is vertex E. From vertex E, you can reach vertex C in 4 + 4 units of distance, which is more than the current value of D(C), 1. This route is not considered as it is clearly proven to be a longer route. With that, we’re done with vertex E.

Now, the process continues to its next iteration and we extract the minimum element from the priority queue. The minimum element in the priority queue is vertex F. You cannot go to any other vertex from vertex F, so, we’re done with vertex F.

With the removal of vertex F, our priority queue becomes empty. So, our algorithm is done…! You can simply return the array D to output the shortest paths.

Having got an idea about the overall working of the Dijkstra’s Algorithm, it’s time to look at the pseudo-code –

In the pseudo-code, G is the input graph and S is the starting vertex. I hope you understand the pseudo-code. If you don’t, feel free to comment your doubts. Now, before we code Dijkstra’s Algorithm, we must first prepare a tool, which is the Priority Queue.

The Priority Queue

The Priority Queue is implemented by a number of data structures such as the Binary Heap, Binomial Heap, Fibonacci Heap, etc. The priority queue in my code is implemented by a simple Binary Min Heap. If you are not aware about the Binary Heap, you can refer to my post on Binary Heaps. Now the functionalities that we need from our priority queue are –

Build Heap – O(|V|) procedure to construct the heap data structure.

Extract Min – O(log |V|) procedure, where we return the top-most element from the Binary Heap and delete it. Finally, we make the necessary changes to the data structure.

Decrease Key – We decrease the priority of an element in the priority queue when we find a shorter path, as known as Relaxation.

If you know the working of the Binary Heap you can code the Priority Queue in about 1-2 hours. Alternatively, you can use C++ STL’s priority queue instead. But you don’t have a “decrease key” method there. So, if you want to use C++ STL’s priority queue, instead of removing elements, you can re-insert the same element with the lowered priority value.

Simple O(|V|2) Implementation

If you choose to implement priority queue simply by using an array, you can achieve the minimum operation in O(N) time. This will give your algorithm a total runtime complexity of O(|V|2). It is the simplest version of Dijkstra’s algorithm. This is the version you are supposed to use if you quickly want to code the Dijkstra’s algorithm for competitive programming, without having to use any fancy data structures. Take a look at the pseudocode again and try to code the algorithm using an array as the priority queue. You can use my code below as a reference –

CJava

Faster O(|E| log |V|) implementation

You can use a binary heap as a priority queue. But remember that to perform the decrease-key operation, you’ll need to know the index of the vertex inside the binary heap array. For that, you’ll need an additional array to store a vertex’s index. Each time any change is made in the binary heap array, corresponding changes must be made in the auxiliary array. Personally, I don’t like this version but I’ll put my code below so that you can use it as a reference.
If you want to do the same thing in C++, you can use a priority queue to reduce a lot. The tweak here is that because you cannot remove a certain vertex from a C++ STL priority queue, we can re-insert it with the new lower priority. This will increase the memory consumption but trust me, its worth it. I have put the codes for both versions below.

C using Binary HeapC++ using STL

The complexity of the above codes is actually O(|V| + |E|) ✗ O(log |V|), which finally makes it O(|E| log |V|). Dijkstra’s Algorithm can be improved by using a Fibonacci Heap as a Priority Queue, where the complexity reduces to O(|V| log |V| + |E|). Because the Fibonacci Heap takes constant time for Decrease Key operation. But the Fibonacci Heap is an incredibly advanced and difficult data structure to code. We’ll talk about that implementation later.

This is the Dijkstra’s Algorithm. If you don’t understand anything or if you have any doubts. Feel free to comment them. I really hope my post has helped you in understanding the Dijkstra’s Algorithm. If it did, let me know by commenting. I tried my best to keep it as simple as possible. Keep practising and… Happy Coding…! 🙂

I am not able to understand the use of hashTable[] properly.
Yes I get that it is for storing the location of vertex i, but the thing which is confusing me is, in the starting at the time of scanning the nodes, the new node’s ‘label’ is the ‘value’ of the hashTable[] while ‘i’ is the ‘key’ and in the heapify procedure, the ‘label’ is being used as the ‘key’ while ‘i’ to be the ‘value’.
Please give a bit detailed explanation for this plus what is the real necessity of taking this hashTable[].
One more thing, are you assuming ‘labels’ to be same as the ‘index’ number at which the node is being stored initially in minHeap[]..i.e from 1 to n only ?
If yes, then what if the labeling is not done in this particular order… what will happen then ?

I completely understand the use of hashTable[] after thinking precisely on its working so you don’t need to explain that part now 😀
But still there is a thing which is a bit problematic to me, i.e If the vertices scanned do not range from [1…vertices] (and have value larger than ‘vertices’) , in that case would we not be able to store the location of the vertex in hashTable[]..?
In my opinion we won’t be able to do that.
Please correct me if I am wrong, and if if not then what would be the solution of that case..?

And Very Thank You for the post 😀 😀
Surely it is the tough algorithm and takes time in the starting to understand each section, but your explanation made it really very helpful to get and code it properly.
Thanks again 🙂

Sorry for the late reply…. Well firstly, even when u have an STL priority queue, you will need a hash table to map the vertices to their indices on the priority queue and secondly, we are using an operation called DecreaseKey() where we lower the priority of an element in the priority queue, and then bubble it up to ensure the structure stays as a heap… Now we don’t have any specific STL function to do that bubbling up for you… So you would have to implement that one function DecreaseKey() by yourself and the rest is taken care by STL… And the complexities stay same as the priority queue in STL is internally a vector or a dequeue which performs similar to an array… 🙂

I did not change the priority of element I just kept on inserting them. Since we take minimum priority element each time we will get correct priority of an element. Although space complexity increases a bit in this approach because of multiple copies of same element, I think it will give correct answer.
Here is my implementation :- http://ideone.com/ugfGvV

The input format is… The first line has 2 integers N and E, the number of nodes and the edges in the graph. This is followed by E number of lines, each consisting of 3 integers, U, V, W… This represents an edge between U and V of weight W. This followed by the last line of the input which is a single integer which is the starting vertex of the algorithm.This is the input file I used to test my code for large input. It is a bi-directional graph where each edge has different weight. The vertices in the input are 1-indexed… 🙂

Hi @Perez..! It was silly of me that I didn’t include that in a Single Source Shortest Path Algorithm…! 😛 … Thanks to remind me..! 🙂 … I have updated the code to print the shortest path also. In order to get the shortest path, we need more information… We need to know the parent node of each node in the shortest path.. That is… The vertex via which we had to come, to get the shortest path. To store this additional information, I convert the shortestDistances array to an array of pairs. The elements of the pair store –

pair.first – Stores the shortest distance from the start vertex to that vertex.

pair.second – Stores the parent vertex in the shortest path.

Then I have made slight changes to the algorithm to get the parents. The changes are in lines 188, 193, 231. They don’t affect the performance of the algorithm. Once you get the information about the parents… To compute the shortest path from E → A, all you have to do is to recursively look at each vertex’s parent. That is exactly what PrintShortestPath() function does. Just give it a little thought, I’m sure you’ll understand it..! I will update my post to include the required explanation in a few days… Feel free to ask if you have any more doubts..! 😀